如何告訴用戶在Python中還有多少次嘗試

Question

我正在嘗試在python中制作隨機數字游戲，其中計算機必須生成1到20之間的數字，而您必須猜測它。 我將猜測的數量限制為6個。當用戶猜錯時，如何打印用戶留下的猜測數量？ 這是我的代碼：

import random

attempts = 0

name = input("What is your name? ")
random = random.randint(1, 20)
print(name + ",","I'm thinking of a number between 1 and 20, What is it?")

while attempts < 6:
    number = int(input("Type your guess: "))
    attempts = attempts + 1
    int(print(attempts,"attemps left")) #This is the code to tell the user how many attempts left
    if number < random:
        print("Too low. Try something higher")
    if number > random:
        print("Too high. Try something lower")
    if number == random:
        break
if number == random:
    if attempts <= 3:
        print("Well done,",name + "! It took you only",attempts,"attempts")
    if attempts >= 4:
        print("Well done,",name + "! It took you",attempts,"attempts. Athough, next time try to get three attempts or lower")
if number != random:
    print("Sorry. All your attempts have been used up. The number I was thinking of was",random)

謝謝，非常感謝您的幫助！

Answer 1

print('attempts left: ', 6 - attempts)

Answer 2

print(6 - attempts, "attempts left")

Answer 3

您的attempts變量將計算使用的嘗試次數。 由於限制是6，因此6 - attempts是剩余的嘗試次數：

print(6 - attempts, "attempts left")

（無需將其包裝為int調用。我不知道您為什么這樣做。）

順便說一句，始終將最大嘗試次數寫為6可能會使6含義模糊不清，並且如果您想將限制更改為例如7，則很難找到所有需要更改的位置。可能值得用描述性名稱：

max_attempts = 6
...
while attempts < max_attempts:
    ...
    print(max_attempts - attempts, "attempts left")

Answer 4

我將提出四個建議，以使您的代碼更簡潔一些：

1. 剔除“幻數” 6並從中倒數而不是增加；
2. 使用for而不是while ，因此您不必手動增加/減少guesses的數量，而使用else確定循環是否break （即，沒有猜測）。
3. 使用if: elif: else:而不是if 和
4. 使用str.format 。

這將使代碼類似於：

attempts = 6
for attempt in range(attempts, 0, -1):
    print("You have {0} attempts left.".format(attempt))
    number = int(input(...))
    if number < random:
        # too low
    elif number > random:
        # too high
    else:
        if attempt > (attempts // 2):
            # great
        else:
            # OK
        break
else:
    # out of guesses